Differentiating Logarithmic Functions with Base e
Add yours
Key Questions

# 1/x # is the derivative of ln(x).
Let us derive the above derivative by Implicit Differentiation.
By rewriting in terms of an exponential function,
#y=lnx Rightarrow e^y =x# by implicitly differentiating with respect to
#x# ,#Rightarrow e^ycdot y'=1# by dividing by
#e^y# ,#Rightarrow y'=1/e^y=1/x# .Hence,
#(lnx)'=1/x# .
I hope that this was helpful.

By Quotient Rule,
#y'={1/x cdot xlnx cdot 1}/{x^2}={1lnx}/{x^2}# This problem can also be solved by the Product Rule
#y'=f'(x)g(x)+f(x)g(x)# The original function can also be rewritten using negative exponents.
#f(x)=ln(x)/x=ln(x)*x^1# #f'(x)=1/x*x^1+ln(x)*1x^2# #f'(x)=1/x*1/x+ln(x)*1/x^2# #f'(x)=1/x^2ln(x)/x^2# #f'(x)=(1ln(x))/x^2# 
The answer would be
#f'(x) = 1/g(x)*g'(x)# or it can be written as#f'(x)=(g'(x))/g(x)# .To solve this derivative you will need to follow the chain rule which states:
#F(x)=f(g(x))# then#F'(x)=f'(g(x))*g'(x)# Or without the equation, it the derivative of the outside(without changing the inside), times the derivative of the outside.
The derivative of
#h(x) = ln(x)# is#h'(x) = 1/x# .
Questions





Doublecheck the answer

Doublecheck the answer

